/*
 * If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
 * Find the sum of all the multiples of 3 or 5 below 1000.
 */
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int i,sum=0;
	for(i=1; i<1000; i++)
	{
		if(i%3==0 || i%5==0)
			sum+=i;
	}
	printf("%d\n", sum);
	return 0;
}
